Download Chromosomal Inheritance

Chi-square Statistical Analysis


You and a sibling flip a coin to see who has to take out the trash. Your
sibling grows skeptical of the legitimacy of your coin because “tails” always
seems to win.
You decide to test it out by flipping the coin 20 times.
Poss. Outcomes Obs. Exp.
Tails
Heads
13 10
7 10
c2 value
(o-e)2/e
.9
.9
1.8
Chi-square Statistical Analysis
Determines the probability that observed data could result from
expected conditions
1. For each phenotype, calculate
(Observed – Expected)2 / (Expected)
2. Add up your figures. This is the chi-square value
2
c =
S
(o-e)2
e
3. Degree of freedom (df) = (# of phenotypes – 1)
4. Find the probability in the table.
Fruit Fly Eye Color
P: Red-eyed female x White-eyed male
 F1: 100 Red-eyed flies
 F2: 75 Red-eyed, 25 white-eyed
Write out Punnett squares for both crosses.

100
75
25
w+ = red
w = white
Fruit Fly Eye Color



P: Red-eyed female x White-eyed male
50 females, 25 males
F1: 100 Red-eyed flies
F2: 75 Red-eyed, 25 white-eyed
25 males
75
25
Chi-square Statistical Analysis


Our F1 cross produces 100 offspring
If we assume eye color and gender are unlinked,
w+ = red
w = white
 Then we expect…
 But we got…


Only a 0.000027%
probability
Therefore, eye
color and gender
are linked
w+w ♀ x w+w ♂
Calc.
Observed
Expected
Phenotype
4.2
50
37.5
Red ♀
12.5
0
12.5
White ♀
4.2
25
37.5
Red ♂
12.5
25
12.5
White ♂
33
c2 value
X-Linkage
 The
Homozygous
Dominant
Xw+Xw+
gene with the
white-eyed mutation is
on the X-chromosome
Hemizygous
Recessive
XwY
Practice Problems
1.
2.
Colorblindness is due to a recessive x-linked
allele. What are the chances of a normal male
and a carrier female having a colorblind son as
their first child?
Why are males more likely than females to
have recessive x-linked traits?
Practice Problems
In sesame plants, the one-pod condition, (A) is dominant to the 3pod condition (a), and normal leaf (B) is dominant to wrinkled leaf
(b). An AaBb plant is testcrossed to produce the following
offspring:
What about… What about…
11
one-pod, normal
110
18
12
one-pod, wrinkled
120
2
7
three-pod, normal
70
3
10
three-pod, wrinkled
100?
17?
Is this data likely if this is a case of independent assortment?
A
AaBb
A
B
a
b
AaBb
A
B
a
b
a
A
A
a
a
B
b
B
B
b
b
A
a
A
A
a
a
b
B
b
b
B
B
A
a
A
A
OR
B
b
B
B
a
b
Gametes
AB
ab
Ab
aB
a
Parental
b
AB
ab
A
a
A
A
a
a
B
b
B
b
B
b
Recomb.
Ab
aB
Earlobes & Toes (Independent Assortment)
P:
FFTT x fftt
all FfTt
(free earloes and 2nd toe longer)
Testcross of F1 individuals
FfTt x fftt:
¼ Free, 2nd toe
¼ Free, great toe
¼ attached, 2nd toe
¼ attached, great toe
Earlobes & Toes (Linked)



P: FFTT x fftt
F1: FfTt
F1 Testcross

F
f
f
f
T
t
t
t
FfTt x fftt
Independent
Assortment
¼ Free, 2nd
¼ Free, Grt
¼ Att, 2nd
¼ Att, Grt
F
f
f
f
T
t
t
t
½ Free,
2nd Toe
½ Att,
Grt Toe
Earlobes & Toes (Linked + Recombination)
F
f
f
f
Tt
Tt
t
t
F
f
f
f
F
f
f
f
T
t
t
t
t
t
t
T
18
17
Parental Offspring
FfTt (free, 2nd)
fftt (attached, great)
2
3
Recombinant Offspring
Fftt (free, great)
ffTt (attached, 2nd)
Recombination frequency = (# of recomb. offspring)
(total offspring)
Genetic Recombination


Genes farther apart are more likely to cross
over
Greater distance  greater RF

Enables us to map genes on a chromosome
Genetic Recombination





b/c
= 9% (9 map units or
centimorgans)
v/c
= 9.5% (9.5 mu)
b/v
= 17% (17 mu)
According to our map, b/v = 18.5 mu
(not 17)
Double crossover leads to lower than
expected recombinant frequencies
v
9.5mu
18.5mu
OR
c
b
v
9mu
9.5mu
Review







Construct a linkage map given the following
information:
a-b  7.2 %
b-c  3.5 %
a-c  3.9 %
b-d  9.6 %
c-d  6.3 %
How many recombinant organisms would you
expect to find out of 1000 offspring from a cross
between AaDd and aadd organisms?